• # question_answer ) The angular velocities of three bodies in simple harmonic motion are ${{\omega }_{1}},{{\omega }_{2}},{{\omega }_{3}}$with respective amplitudes as${{A}_{1}},{{A}_{2}},{{A}_{3}}$If all the three bodies have the same mass and maximum speeds, then: A) ${{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}$   B) ${{A}_{1}}\omega _{1}^{2}={{A}_{2}}\omega _{2}^{2}={{A}_{3}}\omega _{3}^{2}$ C) $A_{1}^{2}{{\omega }_{1}}=A_{2}^{2}{{\omega }_{2}}=A_{3}^{2}{{\omega }_{3}}$D) $A_{1}^{2}\sqrt{{{\omega }_{1}}}=A_{2}^{2}\sqrt{{{\omega }_{2}}}=A_{3}^{2}\sqrt{{{\omega }_{3}}}$

$y=A\sin \omega t$ $v=\frac{dy}{dt}=A\omega \cos \omega t$ $\therefore$Maximum speed $=A\omega$ Since all the bodies have same maximum speeds, so${{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}={{A}_{3}}{{\omega }_{3}}$ Hence, the correction option is [a].